IPhO 2019 · Theory problem 1

Zero-length springs and slinky coils

Download PDFIPhO 2019

A zero effective length spring (ZLS) is a spring for which the force is proportional to the spring's length, \(F=kL\) for \(L>L_0\) where \(L_0\) is the minimal length of the spring as well as its unstretched length. Figure 1 shows the relation between the force \(F\) and the spring length \(L\), for a ZLS, where the slope of the line is the spring constant \(k\).

Figure 1

Figure 1: the relation between the force \(F\) and the spring length \(L\)

A ZLS is useful in seismography and allows very accurate measurement of changes in the gravitational acceleration \(g\). Here, we shall consider a homogenous ZLS, whose weight \(Mg\) exceeds \(kL_0\). We define a corresponding dimensionless ratio, \(\alpha=kL_0/Mg<1\) to characterize the relative softness of the spring. The toy known as “slinky” may be (but not necessarily) such a ZLS.

Part A: Statics (3.0 points)

A.1 Consider a segment of length \(\Delta\ell\) of the unstretched ZLS spring which is then stretched by a force \(F\), under weightless conditions. What is the length \(\Delta y\) of this segment as a function of \(F\), \(\Delta\ell\) and the parameters of the spring? 0.5pt
A.2 For a segment of length \(\Delta\ell\), calculate the work \(\Delta W\) required to stretch it from its original length \(\Delta\ell\) to a length \(\Delta y\). 0.5pt

Throughout this question, we will denote a point on the spring by its distance \(0\leq\ell\leq L_0\) from the bottom of the spring when it is unstretched. In particular, for every point on the spring, \(\ell\) remains unchanged as the spring stretches.

A.3 Suppose that we hang the spring by its top end, so that it stretches under its own weight. What is the total length \(H\) of the suspended spring in equilibrium? Express your answers in terms of \(L_0\) and \(\alpha\). 2.0pt

Part B: Dynamics (5.5 points)

Experiments show that when the spring is hung at rest and then released, it gradually contracts from the top, while the lower part remains stationary (see Figure 2). As time advances, the contracting part moves as a solid chunk and accumulates additional turns of the spring, while the stationary part becomes shorter. Every point on the spring begins to move only when the moving part reaches it. The bottom end of the spring starts moving only when the spring is fully collapsed and reaches its unstretched length \(L_0\). After that, the contracted spring continues falling straight downwards, without tumbling, as a rigid body under the influence of gravity.

Figure 2

Figure 2: Left: a sequence of pictures taken during the free fall of slinky. Right: the moving part I and the stationary part II during the free fall of the spring.

In the remaining parts of the question, you are asked to base your solution on this described model. You may neglect air resistance, but you are not allowed to neglect \(L_0\).

B.1 Calculate the time \(t_c\) it takes from the moment the spring is released, until it fully collapses back to its minimal length \(L_0\). Express your answer in terms of \(L_0\), \(g\) and \(\alpha\). Compute the numerical value of \(t_c\) for a spring with \(k=1.02\ \mathrm{N/m}\), \(L_0=0.055\ \mathrm{m}\) and \(M=0.201\ \mathrm{kg}\), while taking \(g\) to be \(9.80\ \mathrm{m/s^2}\). 2.5pt
B.2 In this task \(\ell\) is used to denote the coordinate of the boundary between parts I (in figure 2, the moving part) and II (the stationary part). At a certain moment, while a stationary part still exists its mass is \(m(\ell)=\frac{\ell}{L_0}M\), and the moving part moves with uniform instantaneous velocity \(v_I(\ell)\). Show that at this moment (while there exists a stationary part) the velocity of the moving part is \(v_I(\ell)=\sqrt{A\ell+B}\). Express the constants \(A\) and \(B\) in terms of \(L_0\), \(g\) and \(\alpha\). 2.5pt
B.3 Based on B.2, find the minimum speed \(v_{\min}\) of the moving part of the spring in the course of its motion, after its release and before it hits the ground. Express your answer in terms of \(L_0\), \(\alpha\), \(A\) and \(B\). 0.5pt

Part C: Energetics (1.5 points)

C.1 Calculate the amount of mechanical energy \(Q\) that was lost by generating heat, from the moment the spring is released until just before the spring hits the ground. Express your answer in terms of \(L_0\), \(M\), \(g\) and \(\alpha\). 1.5pt

Official English text of the problem.

Answers

  • A.1 \[ \boxed{\Delta y=\Delta\ell\max\!\left(1,\frac{F}{kL_0}\right)} \]

  • A.2, for \(\Delta y\geq\Delta\ell\): \[ \boxed{\Delta W=\frac{kL_0}{2\Delta\ell} \left[(\Delta y)^2-(\Delta\ell)^2\right]} \]

  • A.3 \[ \boxed{H=\frac{L_0}{2\alpha}(1+\alpha^2)} \]

  • B.1 \[ \boxed{t_c=\sqrt{\frac{L_0(1-\alpha)^3}{3\alpha g}}} \qquad \boxed{t_c\approx0.245\ \mathrm{s}} \]

  • B.2, taking downward velocity as positive: \[ \boxed{A=\frac{2g}{3\alpha},\qquad B=\frac{gL_0(1-3\alpha)}{3\alpha}} \] The collapse proceeds through \(\alpha L_0\leq\ell<L_0\).

  • B.3 \[ \boxed{v_{\min}=(1-\alpha)\sqrt{A\alpha L_0+B}} \] This occurs immediately after the moving part collides with the initially compact bottom portion.

  • C.1 \[ \boxed{Q=\frac{MgL_0}{6\alpha}(1-\alpha)^2(1+2\alpha)} \]

Solution

A.1

Under weightless conditions the tension is uniform. Since the spring is homogeneous, every segment has the same ratio of stretched length to unstretched length.

When \(F>kL_0\), the whole spring has length \(L=F/k\), so \[ \frac{\Delta y}{\Delta\ell}=\frac{L}{L_0} =\frac{F}{kL_0}. \] Equivalently, the segment has spring constant \[ k_{\mathrm{seg}}=\frac{kL_0}{\Delta\ell}. \]

For \(0\leq F\leq kL_0\), the segment remains at its minimum length. Thus \[ \boxed{ \Delta y= \begin{cases} \Delta\ell,&0\leq F\leq kL_0,\\[2mm] \dfrac{F\Delta\ell}{kL_0},&F>kL_0. \end{cases}} \]

A.2

Increasing the tension up to \(kL_0\) produces no displacement and therefore requires no mechanical work. During subsequent stretching, at instantaneous segment length \(s\), \[ F(s)=\frac{kL_0}{\Delta\ell}s. \] The work required for quasistatic stretching is \[ \Delta W=\int_{\Delta\ell}^{\Delta y}F(s)\,ds =\frac{kL_0}{\Delta\ell} \int_{\Delta\ell}^{\Delta y}s\,ds. \] Hence \[ \boxed{\Delta W= \frac{kL_0}{2\Delta\ell} \left[(\Delta y)^2-(\Delta\ell)^2\right]}. \] This is also the increase in elastic potential energy.

A.3

The mass per unit unstretched length is \[ \mu=\frac{M}{L_0}. \] At material coordinate \(\ell\), the tension supports all the spring below that point: \[ T(\ell)=\mu g\ell=\frac{Mg}{L_0}\ell. \]

Stretching occurs only where \(T(\ell)>kL_0\). The boundary between the compact and stretched portions is therefore \[ \frac{Mg}{L_0}\ell_*=kL_0, \qquad \boxed{\ell_*=\alpha L_0}. \] Thus the lowest fraction \(\alpha\) of the spring remains unstretched.

Let \(y_0(\ell)\) be the equilibrium height above the bottom end. Applying A.1 locally, \[ \frac{dy_0}{d\ell}= \begin{cases} 1,&0\leq\ell\leq\alpha L_0,\\[1mm] \dfrac{\ell}{\alpha L_0},& \alpha L_0\leq\ell\leq L_0. \end{cases} \] Integrating with \(y_0(0)=0\) gives \[ y_0(\ell)= \begin{cases} \ell,&0\leq\ell\leq\alpha L_0,\\[2mm] \dfrac{\ell^2}{2\alpha L_0}+\dfrac{\alpha L_0}{2}, &\alpha L_0\leq\ell\leq L_0. \end{cases} \] In particular, \[ \boxed{H=y_0(L_0)=\frac{L_0}{2\alpha}(1+\alpha^2)}. \]

B.1

After release, gravity is the only external force. Consequently, the center of mass falls with acceleration \(g\), even though internal collisions dissipate energy.

The initial center-of-mass height is \[ \begin{aligned} y_{\mathrm{cm},0} &=\frac{1}{L_0}\int_0^{L_0}y_0(\ell)\,d\ell\\ &=\frac{1}{L_0}\left[ \int_0^{\alpha L_0}\ell\,d\ell+ \int_{\alpha L_0}^{L_0} \left(\frac{\ell^2}{2\alpha L_0} +\frac{\alpha L_0}{2}\right)d\ell \right]\\ &=\frac{L_0}{6\alpha}(1+3\alpha^2-\alpha^3). \end{aligned} \]

At the instant of complete collapse, the bottom end has not yet moved. The spring occupies the interval from \(y=0\) to \(y=L_0\), so \[ y_{\mathrm{cm},c}=\frac{L_0}{2}. \] The center of mass has therefore fallen through \[ \begin{aligned} y_{\mathrm{cm},0}-y_{\mathrm{cm},c} &=\frac{L_0}{6\alpha} (1-3\alpha+3\alpha^2-\alpha^3)\\ &=\frac{L_0}{6\alpha}(1-\alpha)^3. \end{aligned} \] Since it starts from rest, \[ \frac12gt_c^2=\frac{L_0}{6\alpha}(1-\alpha)^3, \] and hence \[ \boxed{t_c=\sqrt{\frac{L_0(1-\alpha)^3}{3\alpha g}}}. \]

For the supplied data, \[ \alpha=\frac{1.02(0.055)}{0.201(9.80)} =0.0284800. \] It is convenient to use the equivalent expression \[ t_c=\sqrt{\frac{M}{3k}}(1-\alpha)^{3/2}. \] Numerically, \[ \boxed{t_c=0.2454228\ldots\ \mathrm{s}\approx0.245\ \mathrm{s}}. \]

B.2

Take downward velocities as positive. Introduce \[ x=L_0-\ell,\qquad D=(1-\alpha)L_0. \] Here \(x\) is the unstretched length already incorporated into the moving compact chunk. Its mass and actual length are \[ m_I=\frac{M}{L_0}x, \qquad L_I=x. \]

Because the remaining spring is stationary, the total downward momentum is \(m_Iv_I\). Integrating the external force \(Mg\) from release gives \[ \frac{M}{L_0}xv_I=Mgt, \] or \[ \boxed{xv_I=L_0gt}. \] This momentum equation includes the effect of the moving chunk gaining mass.

The lower boundary of the moving chunk is at height \(y_0(\ell)\). Its top is consequently at \[ Y_{\mathrm{top}}=y_0(\ell)+L_0-\ell. \] While the boundary is in the initially stretched portion, \[ \begin{aligned} v_I &=-\frac{dY_{\mathrm{top}}}{dt}\\ &=-\left(\frac{\ell}{\alpha L_0}-1\right)\dot\ell\\ &=\frac{D-x}{\alpha L_0}\dot x. \end{aligned} \] Combining this with \(xv_I=L_0gt\), \[ x(D-x)\dot x=\alpha L_0^2gt. \] Integrating from \(x=0,\ t=0\), \[ \frac{Dx^2}{2}-\frac{x^3}{3} =\frac{\alpha L_0^2gt^2}{2}. \] Therefore \[ t^2=\frac{x^2}{\alpha L_0^2g} \left(D-\frac{2x}{3}\right). \] Substitution into the momentum relation gives \[ \begin{aligned} v_I^2 &=\frac{L_0^2g^2t^2}{x^2}\\ &=\frac{g}{\alpha}\left(D-\frac{2x}{3}\right)\\ &=\frac{g}{\alpha} \left[(1-\alpha)L_0-\frac23(L_0-\ell)\right]\\ &=\frac{2g}{3\alpha}\ell +\frac{gL_0(1-3\alpha)}{3\alpha}. \end{aligned} \] Thus \[ \boxed{v_I(\ell)=\sqrt{A\ell+B}}, \qquad \boxed{A=\frac{2g}{3\alpha},\quad B=\frac{gL_0(1-3\alpha)}{3\alpha}}. \]

The moving boundary reaches the initially compact bottom portion at \(\ell=\alpha L_0\). Hence this continuous contraction stage covers \[ \alpha L_0\leq\ell<L_0. \]

B.3

Since \(A>0\), the moving chunk slows down as \(\ell\) decreases. Just before it reaches the initially compact bottom portion, its speed is \[ v_-=\sqrt{A\alpha L_0+B}. \]

At this instant, the moving mass is \((1-\alpha)M\), while the compact bottom mass \(\alpha M\) is still stationary. The spring has reached its minimum total length, so these portions join in a perfectly inelastic collision, as required by the stipulated rigid-body motion afterward.

The gravitational impulse during this instantaneous collision is negligible. Conservation of momentum gives \[ (1-\alpha)Mv_-=Mv_c, \] and therefore \[ v_c=(1-\alpha)\sqrt{A\alpha L_0+B}. \] This is smaller than \(v_-\). Subsequently the entire spring accelerates downward at \(g\), so its speed increases. Consequently, \[ \boxed{v_{\min}=(1-\alpha)\sqrt{A\alpha L_0+B}}. \]

As a check, \[ A\alpha L_0+B=\frac{gL_0(1-\alpha)}{3\alpha}, \] so \[ v_{\min} =\sqrt{\frac{gL_0(1-\alpha)^3}{3\alpha}} =gt_c, \] as required by the total momentum at complete collapse.

The exactly stationary state at release is excluded by “after its release.” In the prescribed continuum model, the moving chunk has vanishing mass as \(t\to0^+\), allowing a finite limiting speed while total momentum tends to zero.

C.1

Choose zero elastic potential energy for the spring at its minimum length. Applying A.2 to an infinitesimal segment gives \[ dU_{\mathrm{el}} =\frac{kL_0}{2} \left[\left(\frac{dy_0}{d\ell}\right)^2-1\right]d\ell. \] The initially compact bottom portion contributes zero. Thus \[ \begin{aligned} U_{\mathrm{el},0} &=\frac{kL_0}{2} \int_{\alpha L_0}^{L_0} \left[\frac{\ell^2}{\alpha^2L_0^2}-1\right]d\ell\\ &=\frac{kL_0^2}{6\alpha^2} (1-3\alpha^2+2\alpha^3). \end{aligned} \] Using \(kL_0=\alpha Mg\), \[ U_{\mathrm{el},0} =\frac{MgL_0}{6\alpha} (1-3\alpha^2+2\alpha^3). \]

Immediately after complete collapse, the elastic energy is zero and the entire spring translates with speed \(v_c=gt_c\). The dissipated energy up to this moment is \[ Q=U_{\mathrm{el},0} +Mg(y_{\mathrm{cm},0}-y_{\mathrm{cm},c}) -\frac12Mv_c^2. \] However, free fall of the center of mass gives \[ Mg(y_{\mathrm{cm},0}-y_{\mathrm{cm},c}) =\frac12Mg^2t_c^2 =\frac12Mv_c^2. \] These two terms cancel, leaving \[ Q=U_{\mathrm{el},0}. \]

After collapse, the rigid spring falls without air resistance or tumbling, so no further mechanical energy is lost before impact. Factoring the result, \[ \boxed{ Q=\frac{MgL_0}{6\alpha} (1-3\alpha^2+2\alpha^3) =\frac{MgL_0}{6\alpha}(1-\alpha)^2(1+2\alpha) }. \]

The solution was checked against the official solution.